First slide
The steps in collision

Body A of mass 4 m moving with speed u collides with another body B of mass 2 m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is


According to conservation of momentum,4mu1=4mv1+2mv2

 2u1-v1=v2                          .........(i) From conversation of energy, 12(4m)u12=12(4m)v12+12(2m)v22  2u12-v12=v22                          ........(ii) From (i) and (ii) 2u12-v12=4u1-v12                    ..........(iii)  Now, fraction of loss in kinetic energy for mass 4m ,  ΔKKi=Ki-KfKi=1/2(4m)u12-12(4m)v1212(4m)u12             .........(iv)  Substituting (iii) in (iv), we get ΔKKi=89

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